x⁴ = xy + 2

⇔  x⁴ - xy = 2

  ⇔ x(x³ - y) = 2

th1 \(\left\{{}\begin{matrix}x=2\\x^3-y=1\end{matrix}\right.\) \(\Leftrightarrow\) \(\left\{{}\begin{matrix}x=2\\y=7\end{matrix}\right.\)

th2 \(\left\{{}\begin{matrix}x=1\\x^3-y=2\end{matrix}\right.\) ⇔ \(\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)

th3 \(\left\{{}\begin{matrix}x=-2\\x^3-y=-1\end{matrix}\right.\) \(\Leftrightarrow\) \(\left\{{}\begin{matrix}x=-2\\y=-7\end{matrix}\right.\)

th4 \(\left\{{}\begin{matrix}x=-1\\x^3-y=-2\end{matrix}\right.\) ⇔ \(\left\{{}\begin{matrix}x=-1\\y=1\end{matrix}\right.\)

50² x 18² + 32 + 100 * 32

= 900² + 32 + 100*32

= 810000 + 32 + 3200

= 813232

`B=4x^2-2x+5=4x^2-2.2x.1/2+1/4+19/4=(2x-1/2)^2+19/4`

Vì \((2x-1/2)^2 \ge 0<=>(2x-1/2)^2+19/4 \ge 19/4\)

  Hay \(B \ge 19/4\)

Dấu "`=`" xảy ra `<=>(2x-1/2)^2=0<=>2x-1/2=0<=>x=1/4`

\(x^2+10x+25\)

\(=x^2+2.5x+5^2\)

\(=\left(x+5\right)^2\)

\(x^2-6x+9\)

\(=x^2-2.3x+3^2\)

\(=\left(x-3\right)^2\)

\(x^2-x+\dfrac{1}{4}\)

\(=x^2-2.\dfrac{1}{2}x+\left(\dfrac{1}{2}\right)^2\)

\(=\left(x-\dfrac{1}{2}\right)^2\)

\(x^2y^4-2xy^2+1\)

\(=\left(xy^2\right)^2-2.1xy^2+1^2\)

\(=\left(xy^2-1\right)^2\)

A = 4x² - 12 xy + 9y² 

A = (2x)² - 2.2x.3y + (3y)²

A = (2x - 3y)² 

A(1/2;2/3) = (2. \(\dfrac{1}{2}\) - 3.\(\dfrac{2}{3}\))² = (1-2)² =(-1²) = 1

\(a,=-\left(x^3-3x^2y+3xy^2-y^3\right)\\ =-\left(x-y\right)^3\\ b,=\left(5-x\right)^3\)

Bài 1:

b) 125-75x+15x²-x³= (5-x)³

Bài 2:

b) x³-12x²+48x-64=0

⇔ (x-4)³=0

⇔x-4=0

⇔x= 4

`(a-b+c)^2`

`=[(a-b)+c]^2`

`=(a-b)^2+2c(a-b)+c^2`

`=a^2-2ab+b^2+2ac-2bc+c^2`

`=a^2+b^2+c^2-2ab+2ac-2bc`

a^2-2ab+2ac+b^2-bc+c^2-cb